Figure A-7.5(a), a graph of $f\left(x\right)\=4 \sin \left(x\right)\+7 \cos \left(x\right)-8$, suggests that in the interval $\left[0\,2 \mathrm{\π}\right]$, the given equation has two solutions, both of which appear to be in the first quadrant.

Two different functions appear in the equation, but before the trig identity ${\sin }^2\left(x\right)\=1-{\cos }^2\left(x\right)$ can be used to convert the equation to one quadratic in $\cos \left(x\right)$, the sine and cosine terms must appear on different sides of the equation, both sides of which must then be  squared.

However, this last step can, and does, introduce extraneous roots!

Isolate $\sin \left(x\right)$ to obtain $\sin \left(x\right)\=-\frac{7}{4}\cos \left(x\right)\+2$, then square both sides to obtain ${\sin \left(x\right)}^2\={\left(-\frac{7}{4}\cos \left(x\right)\+2\right)}^2$. Replace ${\sin }^2\left(x\right)$ with $1-{\cos }^2\left(x\right)$ and simplify, to obtain $\frac{65}{16}{\cos \left(x\right)}^2-7\cos \left(x\right)\+3\=0$.

The transformed equation factors to $\frac{1}{16}\left(5\cos \left(x\right)-4\right)\left(13\cos \left(x\right)-12\right)\=0$.

By the Zero Principle, either or both factors must themselves be zero, so the following two equations must be solved.

The solutions of the first are  $x\=\arccos \left(4\/5\right)\doteq 0.644$  and $x\=2 \mathrm{\π}-\arccos \left(4\/5\right)\doteq 5.640$, the second of which is extraneous.

The solutions of the second are $x\=\arccos \left(12\/13\right)\doteq 0.395$ and $x\=2 \mathrm{\π}-\arccos \left(12\/13\right)\doteq 5.888$, the second of which is extraneous.

Tools≻Load Package: Student Calculus 1

Control-drag the equation; press the Enter key.

Context Panel: Solve≻Isolate Expression for≻$\sin \left(x\right)$ 

Context Panel: Manipulate Equation
Square both sides.
Press Return Steps button.

Expression palette: Evaluation template
Evaluate at ${\sin }^2\left(x\right)\=1-{\cos }^2\left(x\right)$ 
Press the Enter key.

Context Panel: Move to Left

Context Panel: Left-hand Side

Context Panel: Factor

${\sin \left(x\right)}^2\={\left(-\frac{7}{4}\cos \left(x\right)\+2\right)}^2$

Control-drag the first factor; equate to zero.
Press the Enter key.

Context Panel:
Student Calculus1≻Solve≻Find Roots
Enter bounds as per Figure A-7.5(b)

Context Panel: Approximate≻5 (digits)

Control-drag the second factor; equate to zero.
Press the Enter key.

Context Panel:
Student Calculus1≻Solve≻Find Roots
Enter bounds as per Figure A-7.5(b)

Context Panel: Approximate≻5 (digits)

Tools≻Load Package: Student Calculus 1
(Skip this step if package already loaded.)

Assign the equation to the name $\mathrm{q\_\_1}$.

Apply the Roots command with the appropriate bound on $x$.

Apply the evalf command with 5 as the optional digits parameter.

Apply the isolate command to isolate $\sin \left(x\right)$.

Map the function $x\to x^2$ onto the equation to square both sides.

Use the lhs and rhs command to bring all terms to the left, and apply the simplify command, which favors $\cos \left(x\right)$ over $\sin \left(x\right)$.

Apply the factor command.